Simplify the following expression and state the condition under which the simplification is valid. You can assume that $t \neq 0$. $k = \dfrac{-5}{6t + 27} \times \dfrac{t(2t + 9)}{-10} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ -5 \times t(2t + 9) } { (6t + 27) \times -10 } $ $ k = \dfrac {-5 \times t(2t + 9)} {-10 \times 3(2t + 9)} $ $ k = \dfrac{-5t(2t + 9)}{-30(2t + 9)} $ We can cancel the $2t + 9$ so long as $2t + 9 \neq 0$ Therefore $t \neq -\dfrac{9}{2}$ $k = \dfrac{-5t \cancel{(2t + 9})}{-30 \cancel{(2t + 9)}} = -\dfrac{5t}{-30} = \dfrac{t}{6} $